# In a certain candy store, 3 pounds of candy and 2 pounds of mints cost $10.80, and 1 pound of candy and 3 pounds of mints cost$5.35. What is the cost per pound of the mints?

Jul 26, 2016

Mints cost $3/4 per pound I have taken you up to this point and given a strong hint about to solve the next bit. #### Explanation: The trick with these question to 'get rid of 1 of the unknowns so that you can solve for the remaining one. Let the cost of candy per pound be $c$Let the cost of mints per pound be $m$Then we have: 3c+2m=$10.80.............................(1)
c+3m=$5.35....................................(2) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Consider equation(2) $\text{ } \textcolor{b r o w n}{c + 3 m = 5.35}$Subtract $\textcolor{b l u e}{3 m}$from both sides $\text{ } \textcolor{b r o w n}{c + 3 m \textcolor{b l u e}{- 3 m} = 5.35 \textcolor{b l u e}{- 3 m}}$But $3 m - 3 m = 0$$\text{ } c + 0 = 5.35 - 3 m$$\text{ } c = 5.35 - 3 m \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Substitute $\left({2}_{a}\right)$into (1) color(brown)(3c+2m=$10.80)color(blue)(" "->" "3(5.35-3m)+2m=10.8

$\text{ } 16.05 - 9 m + 2 m = 10.8$

$\text{ } 16.05 - 7 m = 10.8$

$\text{ } 7 m = 16.05 - 10.8$

$\text{ "7m=5.25 larr" note that 5.25 is } 5 \frac{1}{4}$

$\text{ } m = \frac{1}{{\cancel{7}}^{1}} \times \frac{{\cancel{21}}^{3}}{4} =$

$\text{ } m = \frac{3}{4}$
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$\textcolor{g r e e n}{\text{I will let you solve for the cost of candy}}$

Hint: substitute $m = \frac{3}{4}$ in one of the original equations